Binning of integers with hist() function odd results (P (PR#14048)

On 06-Nov-09 23:30:12, gug at fnal.gov wrote:

Full_Name: Gerald Guglielmo
Version: 2.8.1 (2008-12-22)
OS: OSX Leopard
Submission from: (NULL) (131.225.103.35)

When I attempt to use the hist() function to bin integers the  
behavior
seems
very odd as the bin boundary seems inconsistent across the various
bins. For
some bins the upper boundary includes the next integer value, while  
in
others it
does not. If I add 0.1 to every value, then the hist() binning  
behavior
is what
I would normally expect.

h1<-hist(c(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5))
h1$mids

[1] 1.5 2.5 3.5 4.5

h1$counts

[1] 3 3 4 5

h2<- 
hist(c(1.1,2.1,2.1,3.1,3.1,3.1,4.1,4.1,4.1,4.1,5.1,5.1,5.1,5.1,5.1)
)
h2$mids

[1] 1.5 2.5 3.5 4.5 5.5

h2$counts

[1] 1 2 3 4 5

Naively I would have expected the same distribution of counts in the
two cases, but clearly that is not happening. This is a simple  
example
to illustrate the behavior, originally I noticed this while binning a
large data sample where I had set the breaks=c(0,24,1).

This is the correct intended bahaviour. By default, values which are
exactly on the boundary between two bins are counted in the bin which
is just below the boundary value. Except that the bottom-most break
will count values on it into the bin just above it.

Hence 1,2,2 all go into the [1,2] bin; 3,3,3 into (2,3];
4,4,4,4 into (3,4]; and 5,5,5,5,5 into (4,5]. Hence the counts  
3,3,4,5.

Since you did not set breaks in
 h1<-hist(c(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5)),
they were set using the default method, and you can see what they are
with

 h1$breaks
 [1] 1 2 3 4 5

When you add 0.1 to each value, you push the values on the boundaries
up into the next bin. Now each value is inside its bin, and not on
any boundary. Hence 1.1 is in (1,2]; 2.1,2.1 in (2,3];
3.1,3.1,3.1 in (3,4]; 4.1,4.1,4.1,4.1 in (4,5]; and
5.1,5.1,5.1,5.1,5.1 in (5,6], giving counts 1,2,3,4,5 as you observe.

The default behaviour described above is defined by the default  
options

 include.lowest = TRUE, right = TRUE

where:

include.lowest: logical; if 'TRUE', an 'x[i]' equal to the 'breaks'
         value will be included in the first (or last, for 'right =
         FALSE') bar.  This will be ignored (with a warning) unless
         'breaks' is a vector.

  right: logical; if 'TRUE', the histograms cells are right-closed
         (left open) intervals.

See '?hist'. You can change this behaviour by shanging the options.

Hoping this helps,
Ted.

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E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 07-Nov-09                                       Time: 13:57:07
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